A particle is projected from a horizontal plane with a velocity of `8sqrt2 ms ^(-1)` ms at an angle `theta`. At highest point its velocity is found to be `8 ms^(-1)` Its range will be `(g=10 ms^(-2))`

Options

A. 3.2 m
B. 4.6 m
C. 6.4 m
D. 12.8 m

Explanation

To solve the problem, we need to find the range of a projectile that is launched with an initial velocity of \(8\sqrt{2} \, \text{m/s}\) at an angle \(\theta\). We know that at the highest point of its trajectory, the velocity is \(8 \, \text{m/s}\). We will use the principles of projectile motion to derive the range.

### Step-by-Step Solution:

1. **Understanding the Components of Velocity:**
The initial velocity \(u\) can be broken down into its horizontal and vertical components:
\[ u_x = u \cos \theta \quad \text{and} \quad u_y = u \sin \theta \]
where \(u = 8\sqrt{2} \, \text{m/s}\).

2. **Velocity at the Highest Point:**
At the highest point of the projectile's motion, the vertical component of the velocity \(u_y\) becomes zero. Therefore, the velocity at the highest point is purely horizontal:
\[ v_x = u_x = u \cos \theta \]
Given that \(v_x = 8 \, \text{m/s}\), we can set up the equation:
\[ u \cos \theta = 8 \]
Substituting \(u = 8\sqrt{2}\):
\[ 8\sqrt{2} \cos \theta = 8 \]
Dividing both sides by 8:
\[ \sqrt{2} \cos \theta = 1 \]
Therefore:
\[ \cos \theta = \frac{1}{\sqrt{2}} \implies \theta = 45^\circ \]

3. **Calculating the Range:**
The formula for the range \(R\) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Substituting \(\theta = 45^\circ\) (where \(\sin 90^\circ = 1\)):
\[ R = \frac{(8\sqrt{2})^2 \cdot 1}{g} \]
Calculating \(u^2\):
\[ (8\sqrt{2})^2 = 64 \cdot 2 = 128 \]
Now substituting \(g = 10 \, \text{m/s}^2\):
\[ R = \frac{128}{10} = 12.8 \, \text{m} \]

### Final Answer:
The range of the projectile is \(12.8 \, \text{m}\).

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